The Stacks project

Proof. By Zariski's main theorem we can choose a factorization $f = f' \circ j$ where $j : Y \to Y'$ is an open immersion and $f' : Y' \to X$ is a finite morphism, see More on Morphisms, Lemma 37.43.3. By our construction in Duality for Schemes, Lemma 48.16.2 we have $f^! = j^* \circ a'$ where $a' : D_\mathit{QCoh}(\mathcal{O}_ X) \to D_\mathit{QCoh}(\mathcal{O}_{Y'})$ is the right adjoint to $Rf'_*$ of Duality for Schemes, Lemma 48.3.1. By Duality for Schemes, Lemma 48.11.4 we see that $\Phi (a'(\mathcal{O}_ X)) = R\mathop{\mathcal{H}\! \mathit{om}}\nolimits (f'_*\mathcal{O}_{Y'}, \mathcal{O}_ X)$ in $D_\mathit{QCoh}^+(f'_*\mathcal{O}_{Y'})$. In particular $a'(\mathcal{O}_ X)$ has vanishing cohomology sheaves in degrees $< 0$. The zeroth cohomology sheaf is determined by the isomorphism

as $f'_*\mathcal{O}_{Y'}$-modules via the equivalence of Morphisms, Lemma 29.11.6. Writing $(f')^{-1}U = V' = \mathop{\mathrm{Spec}}(B')$, we obtain

As the zeroth cohomology sheaf of $a'(\mathcal{O}_ X)$ is a quasi-coherent module we find that the restriction to $V$ is given by $\omega _{B/A} = \mathop{\mathrm{Hom}}\nolimits _ A(B', A) \otimes _{B'} B$ as desired.

The statement about restriction maps signifies that the restriction mappings of the quasi-coherent $\mathcal{O}_{Y'}$-module $H^0(a'(\mathcal{O}_ X))$ for opens in $Y'$ agrees with the maps defined in Lemma 49.2.3 for the modules $\omega _{B/A}$ via the isomorphisms given above. This is clear.

Let $f : Y \to X$ be a quasi-finite morphism of schemes of finite type over a Noetherian base $S$ endowed with a dualizing complex $\omega _ S^\bullet$. Consider opens $V \subset Y$ and $U \subset X$ with $f(V) \subset U$ and $V$ and $U$ separated over $S$. Denote $f|_ V : V \to U$ the restriction of $f$. By the discussion above and Duality for Schemes, Lemma 48.20.9 there are canonical isomorphisms

We omit the verification that these isomorphisms glue to a global isomorphism $H^0(f_{new}^!\mathcal{O}_ X) \to \omega _{Y/X}$. $\square$

Proof. The functor $f^!$ is defined in Duality for Schemes, Section 48.16. Since $f$ is finite (and hence proper), we see that $f^!$ is given by the right adjoint to pushforward for $f$. In Duality for Schemes, Section 48.11 we have made this adjoint explicit. In particular, the object $f^!\mathcal{O}_ X$ consists of a single cohomology sheaf placed in degree $0$ and for this sheaf we have

To see this we use also that $f_*\mathcal{O}_ Y$ is finite locally free as $f$ is a finite flat morphism of Noetherian schemes and hence all higher Ext sheaves are zero. Some details omitted. Thus finally

On the other hand, we have $f^!\mathcal{O}_ X = \omega _{Y/X}$ by the identification of Lemma 49.15.1. Thus we now have two elements, namely $\text{Trace}_ f$ and $\tau _{Y/X}$ from Remark 49.4.7 in

and the lemma says these elements are the same.

Let $U = \mathop{\mathrm{Spec}}(A) \subset X$ be an affine open with inverse image $V = \mathop{\mathrm{Spec}}(B) \subset Y$. Since $f$ is finite, we see that $A \to B$ is finite and hence the $\omega _{Y/X}(V) = \mathop{\mathrm{Hom}}\nolimits _ A(B,A)$ by construction and this isomorphism agrees with the identification of $f_*f^!\mathcal{O}_ Y$ with $\mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_ X}(f_*\mathcal{O}_ Y, \mathcal{O}_ X)$ discussed above. Hence the agreement of $\text{Trace}_ f$ and $\tau _{Y/X}$ follows from the fact that $\tau _{B/A} = \text{Trace}_{B/A}$ by Lemma 49.4.3. $\square$